1970 IMO Problems/Problem 1
Contents
[hide]Problem
Let be a point on the side of . Let , and be the inscribed circles of triangles , and . Let , and be the radii of the escribed circles of the same triangles that lie in the angle . Prove that
.
Solution
We use the conventional triangle notations.
Let be the incenter of , and let be its excenter to side . We observe that
,
and likewise,
Simplifying the quotient of these expressions, we obtain the result
.
Thus we wish to prove that
.
But this follows from the fact that the angles and are supplementary.
Solution 2
By similar triangles and the fact that both centers lie on the angle bisector of , we have , where is the semi-perimeter of . Let have sides , and let . After simple computations, we see that the condition, whose equivalent form is is also equivalent to Stewart's Theorem (see Stewart's_theorem or https://en.wikipedia.org/wiki/Stewart's_theorem)
Solution 3
Let be the incenter, and be the excenter relative to C, and let be the points where the incircle and the excircle relative to touch .
The triangles and are similar, so
Let be the incenter, and be the excenter relative to C of the triangle , and be the incenter, and be the excenter relative to C of the triangle ( is not shown on the picture).
To solve the problem, we need to prove that
Applying the law of sines (see Law of Sines or https://en.wikipedia.org/wiki/Law_of_sines) in triangle we get
, or . Since this becomes
Similarly, using the law of sines in triangle and replacing some angles, we get
It follows that
Now, we proceed like in the first solution: Apply the above to triangles and .
We get that and
The desired equality follows from (since and are complementary).
[Solution by pf02, November 2024]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1970 IMO (Problems) • Resources | ||
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